1. The performance equation of a CSTR is ____
a) \(\frac{F_{A_0}}{V} = \frac{(-r_A)}{X_A} \)
b) \(\frac{F_{A_0}}{V} = \frac{X_A}{(r_A)} \)
c) \(\frac{V}{X_A} = \frac{(-r_A)}{F_{A_0}} \)
d) \(\frac{F_{A_0}}{V} = \frac{X_A}{-r_A} \)
Explanation: For a CSTR, Molar flow rate in = Molar flow rate out + accumulation
FA0 = FA + (-rA)V
FA0 = FA0(1- XA) + (-rA)V
Hence, FA0XA = (-rA)V.
2. The performance equation of a PFR is ____
a) \(\frac{F_{A_0}}{dV} = \frac{dX_A}{(-r_A)} \)
b) \(\frac{dV}{F_{A_0}} = \frac{(-r_A)}{dX_A} \)
c) \(\frac{F_{A_0}}{dV} = \frac{(-r_A)}{dX_A} \)
d) \(\frac{F_{A_0}}{dX_A} = \frac{(-r_A)}{dV} \)
Explanation: FA0dXA = (-rA)dV
The reaction is integrated in the entire volume of reactor. The material balance is carried out in a small differential volume dV.
3. If the conversion of a first order liquid phase reaction occurring in a CSTR is 75%, molar feed rate is 5 mol/min, the rate of the reaction is 5 \(\frac{mol}{litre.min} \) then the volume of the reactor (in litre) is ____
a) 0.5
b) 0.75
c) 0.33
d) 0.4
Explanation: For a CSTR, the performance equation is FA0XA = (-rA)V
FA0 is molar feed rate and XA is the fractional conversion
5 × 0.75 = 5 × V
V = 0.75 L
4. If CA is the final concentration and CA0 is the initial concentration, the conversion of a reaction is expressed as ____
a) \(\frac{C_{A0}-C_A}{(C_{A0})} \)
b) \(\frac{C_{A0}-C_A}{(-r_A)}\)V
c) \(\frac{C_{A0}-C_A}{(-r_A)V} \)
d) \(\frac{C_A}{(C_{A0})} \)
Explanation: Conversion in a chemical reaction is the ratio of the amount of reactant consumed to form products to the amount of reactant fed. If the entire reactant is converted to product, then the conversion is 100%.
5. Which of the following is NOT a part of reactor design?
a) Size of the reactor
b) Type of the reactor
c) Flow rates of the reactant and product streams
d) Method of operation
Explanation: In general, most reactors are not a stand-alone unit. They are part of huge plants. This means that the flow rates of reactant and/or product streams depend on a lot of factors and may even be pre-decided. The purpose of reactor design, therefore, is to determine the reactor size and type and the suitable method of operation to meet the flow rate or other requirements.
6. Choose the correct option.
a) Batch reactors are used in laboratories for calorimetric titrations
b) The steady state flow reactor is the most widely used because it needs little supporting equipment
c) Batch reactors are ideal for fast reactions
d) In the steel industry, semi batch reactors are used for large open hearth furnaces
Explanation: Semi batch reactors offer good control on reaction speed and, so, are used in calorimetric titrations and large furnaces. Steady state flow reactors are best suited for fast reactions but their supporting equipment needs are significant.
7. For any type of reactor, the material balance equation is valid.
[Rate of reactant flowing into element of volume] = [Rate of reactant flowing out of element of volume] + [Rate of reactant loss due to chemical reaction within the volume element] + [Rate of accumulation of reactant in the volume element]
Which of the following is false?
a) Two terms disappear in case of a batch reactor
b) Only three terms are non-trivial when we analyze a semi batch reactor
c) One term becomes zero for a steady-state flow reactor
d) There is no flow of reactants in a batch reactor
Explanation: A semi batch reactor represents most general cases where all of the terms in the material balance equation may have to be considered. Whereas, for a batch reactor the first two terms disappear as there is no flow of reactants involved and the accumulation term becomes zero for a steady state flow reactor.
8. Consider the liquid reaction 3A + 2B → 4C
What could be the outlet composition if CAo = 100 mol/L in a batch reactor and the feed is an equimolar mixture of A and B? Assume that 5 mol of C was already present in the reactor initially.
a) XA = 0.2, CB = 86.66, Cc = 31.66
b) XA = 0.4, CB = 76.66, Cc = 58.33
c) XA = 0.6, CB = 60, Cc = 80
d) XA = 0.8, CB = 43.33, Cc = 106.66
Explanation: Taking A as the key reactant and using material balance we get,
CB = CAo*(1 – 2*XA/3)
Cc = CAo*(5/100 + 4*XA/3)
9. Consider the gas reaction A + B → 3C
Which of the following cannot be the outlet composition if FAo = 100 mol/s in a flow reactor with vo = 10 L/s and the feed is an equimolar mixture of A and B? Take the initial conditions as 1 atm, 273K and outlet condition as 2 atm, 300K.
a) XA = 0.2, CB = 13.23, Cc = 5.45
b) XA = 0.4, CB = 9.1, Cc = 18.2
c) XA = 0.6, CB = 5.6, Cc = 25.19
d) XA = 0.8, CB = 2.6, Cc = 31.19
Explanation: Taking A as the key reactant and using material balance and ideal gas equation we get,
CB = [FAo*(1 – XA)/(1 + 0.5*XA)]*[273/150*vo]
Cc = [FAo*(3*XA)/(1 + 0.5*XA)]*[273/150*vo]
10. Percy prepares vegetable curry (liquid) in a pressure cooker. If the raw potatoes are P, the raw tomatoes are T, the raw onions are O, and the cooked meal is M, then 2P + T + O → 3M may represent the cooking process. Sometime during the process, Manu opens the lid and checks the curry to find that the potatoes are half-cooked (take XP = 0.5). Volume of the cooker is 5L and CPo = CTo = COo = 1.5. What can you say about the composition in the cooker at the instant of time when the lid is opened, if initial temperature was 310K?
a) XT = 0.35
b) CO = 1.125
c) T = 380K
d) CM = 2.5
Explanation: If we assume the pressure cooker to be a batch reactor, P, T, O as the reactants and M as the product, we can use material balance to get
XT = CAo*b*XA/(a*CBo)
CO = CAo*(1-0.5*XA)
CM = CAo*(3*XA)
It is a liquid reaction so change in volume can be ignored. Thus we cannot say anything about the temperature based on the given information.